(a) If C(x) is the cost of producing x units of a commodity,...

(a) If $C(x)$ is the cost of producing $x$ units of a commodity, then the average cost per unit is $c(x)=C(x) / x$ . Show that if the average cost is a minimum, then the marginal cost equals the average cost. (b) If $C(x)=16,000+200 x+4 x^{3 / 2},$ in dollars, find (i) the cost, average cost, and marginal cost at a pro- duction level of 1000 units; (ii) the production level that will minimize the average cost; and (iii) the mini- mum average cost.

(a) If $C(x)$ is the cost of producing $x$ units of a commodity, then the average cost per unit is $c(x)=C(x) / x$ . Show that if the average cost is a minimum, then the marginal
cost equals the average cost.
(b) If $C(x)=16,000+200 x+4 x^{3 / 2},$ in dollars, find
(i) the cost, average cost, and marginal cost at a pro-
duction level of 1000 units; (ii) the production level
that will minimize the average cost; and (iii) the mini-
mum average cost.

Key Concepts

Cost Function

A cost function assigns a total cost to producing a certain number of units. It is fundamental in economics and production theory where determining the costs associated with various levels of production helps in planning, budgeting, and decision-making. The overall cost function is usually denoted by C(x), where x is the number of units produced.

Average Cost

The average cost is the cost per unit of production calculated by dividing the total cost by the number of units produced, i.e., c(x) = C(x)/x. It provides an insight into the efficiency of production and helps in determining the scale of operation that minimizes cost per unit.

Marginal Cost

Marginal cost represents the change in total cost resulting from the production of one additional unit of output. It is found by taking the derivative of the cost function with respect to the number of units produced, i.e., C'(x). Marginal cost is a crucial concept in optimization since it indicates how costs change with incremental production adjustments.

Optimization and Derivative in Cost Analysis

Optimization in cost analysis involves finding the production level that minimizes the average cost. Using calculus, this is typically achieved by differentiating the average cost function with respect to the number of units and setting the derivative to zero to find the critical points. It is an essential method in identifying efficient production levels.

Relationship between Marginal and Average Cost at the Minimum Average Cost

When the average cost is minimized, the derivative of the average cost function equals zero. Deriving the average cost function c(x) = C(x)/x using the quotient rule yields an expression that, when set equal to zero, shows that xC'(x) - C(x) = 0. Simplifying this equation leads to C'(x) = C(x)/x, meaning that the marginal cost equals the average cost at the point where the average cost is minimized.

Transcript

00:01 Here we are told that if c of x is a cost of producing x units, then c of x over x is the average cost per unit.

00:11 For part a, we want to show that if the average cost is minimum, then the marginal cost is equal to the average cost.

00:28 So here we just have to set the derivative of the average cost equal to zero and then solve.

00:48 So that is the average cost.

00:51 So when we go to differentiate, we can just use quotient rule.

01:01 And we get that that is x times c prime of x minus c of x all over x squared.

01:16 And then setting that equal to zero, we get that x times c prime of x is equal to c of x.

01:31 And that means that c prime of x is equal to c of x all over x.

01:46 And that's what that's just what we wanted because c prime of x is, of x, that's our marginal cost, and then c of x over x, that's the average cost.

02:03 So that's our solution for part a.

02:14 And then when we get to part b, we're given the cost function 16 ,000 plus 200x, plus four times x to the three halves.

02:24 And we want to find for part one, we want to find the cost, the average cost, and the marginal cost at a thousand units.

02:34 We know that the cost was given to us and that's just c of 1000 and that is approximately 342 ,491 and so that's our cost.

03:03 And then we want the average cost so that's just c of 1000 all over 1 ,000 and that's approximately 342 .49.

03:43 And then next we want the marginal cost.

03:46 And the marginal cost is just c prime of a thousand.

03:58 So when we differentiate our cost function, we get 200 plus three halves times four, which is six times the square root of a thousand.

04:22 And that's approximately equal to 389 .7...

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